Question

directions: type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s).
nathan opened an interest - bearing savings account two years ago. his account balance, in dollars, can be calculated using the expression shown below, where t is time, in years.
$3,600(1.07)^{12t}$
use the given expression to complete the statements below.
the quantity that represents nathans initial deposit is
the quantity that represents the rate at which the account is increasing is, which means the interest rate on the account is %.
the quantity that represents the number of times the account compounds each year is

Explanation:

Step1: Find the initial deposit

The formula for compound - interest is \(A = P(1+\frac{r}{n})^{nt}\), where \(P\) is the principal (initial deposit).
In the given formula \(A = 3600(1.07)^{12t}\), when \(t = 0\) (at the start, time \(t = 0\) years), \(A=3600(1.07)^{0}=3600\). So the initial deposit \(P = 3600\).

Step2: Find the rate of increase

We know that \(A = P(1+\frac{r}{n})^{nt}\). In the formula \(A = 3600(1.07)^{12t}\), if we compare it with \(A = P(1 + r)^{nt}\) (assuming \(n = 1\) for simplicity of rate calculation from the base of the exponent).
Let \(x=1 + r\), here \(x = 1.07\). So \(r=x - 1=0.07\). The rate of increase is \(0.07\) and the interest rate is \(7\%\) (since \(r\times100\%=0.07\times100\% = 7\%\)).

Step3: Find the number of compounding times per year

Comparing \(A = 3600(1.07)^{12t}\) with the compound - interest formula \(A=P(1 + r)^{nt}\), we can see that \(n = 12\).

Answer:

The quantity that represents Nathan's initial deposit is \(3600\). The quantity that represents the rate at which the account is increasing is \(0.07\), which means the interest rate on the account is \(7\%\). The quantity that represents the number of times the account compounds each year is \(12\).